62. Search in Rotated Sorted Array【medium】

62. Search in Rotated Sorted Array【medium】

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Have you met this question in a real interview?

Yes

Example

For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

Challenge

O(logN) time

错误解法：

 1 class Solution {
 2 public:
 3     /*
 4      * @param A: an integer rotated sorted array
 5      * @param target: an integer to be searched
 6      * @return: an integer
 7      */
 8     int search(vector<int> A, int target) {
 9         if (A.size() == 0) {
10             return -1;
11         }
12
13         int start = 0;
14         int end = A.size() - 1;
15
16         while (start + 1 < end) {
17             int mid = start + (end - start) / 2;
18
19             if (A[mid] == target) {
20                 return mid;
21             }
22             else if (A[mid] < target) {
23                 // 5 6 1 2 3 4
24                 if (A[mid] > A[start]) {
25                     start = mid;
26                 }
27                 // 5 6 1 2 3 4
28                 else { // mid < target && mid <= start
29                     end = mid;
30                 }
31             }
32             else if (A[mid] > target) {
33
34                 if (A[mid] < A[end]) { // mid > target && mid < end
35                     end = mid;
36                 }
37                 // 3 4 5 6 1 2 找3
38                 else { // mid > target && mid >= end
39                     start = mid;
40                 }
41             }
42         }
43
44         if (A[start] == target) {
45             return start;
46         }
47
48         if (A[end] == target) {
49             return end;
50         }
51
52         return -1;
53     }
54 };

一开始思考的方向就不对，搞晕了……

解法一：

 1 class Solution {
 2 public:
 3     /*
 4      * @param A: an integer rotated sorted array
 5      * @param target: an integer to be searched
 6      * @return: an integer
 7      */
 8     int search(vector<int> A, int target) {
 9         if (A.size() == 0) {
10             return -1;
11         }
12
13         int start = 0;
14         int end = A.size() - 1;
15
16         while (start + 1 < end) {
17             int mid = start + (end - start) / 2;
18
19             if (A[mid] == target) {
20                 return mid;
21             }
22
23             if (A[mid] >= A[start]) {
24                 if (A[start] <= target && target <= A[mid]) {
25                     end = mid;
26                 }
27                 else {
28                     start = mid;
29                 }
30             }
31             else {
32                 if (A[mid] <= target && target <= A[end]) {
33                     start = mid;
34                 }
35                 else {
36                     end = mid;
37                 }
38             }
39         }
40
41         if (A[start] == target) {
42             return start;
43         }
44
45         if (A[end] == target) {
46             return end;
47         }
48
49         return -1;
50     }
51 };

借鉴网上的一个图，可以清楚的归纳一下思路，那么代码就好写了。

这个图参考了：http://fisherlei.blogspot.com/2013/01/leetcode-search-in-rotated-sorted-array.html

解法二：

 1 class Solution {
 2 public:
 3     int search(int A[], int n, int target) {
 4         return searchRotatedSortedArray(A, 0, n-1, target);
 5     }
 6
 7     int searchRotatedSortedArray(int A[], int start, int end, int target) {
 8         if(start>end) return -1;
 9         int mid = start + (end-start)/2;
10         if(A[mid]==target) return mid;
11
12         if(A[mid]<A[end]) { // right half sorted
13             if(target>A[mid] && target<=A[end])
14                 return searchRotatedSortedArray(A, mid+1, end, target);
15             else
16                 return searchRotatedSortedArray(A, start, mid-1, target);
17         }
18         else {  // left half sorted
19             if(target>=A[start] && target<A[mid])
20                 return searchRotatedSortedArray(A, start, mid-1, target);
21             else
22                 return searchRotatedSortedArray(A, mid+1, end, target);
23         }
24     }
25 };

解法三：

 1 class Solution {
 2 public:
 3     int search(int A[], int n, int target) {
 4         int start = 0, end = n-1;
 5         while(start<=end) {
 6             int mid = start + (end-start)/2;
 7             if(A[mid]==target) return mid;
 8
 9             if(A[mid]<A[end]) { // right half sorted
10                 if(target>A[mid] && target<=A[end])
11                     start = mid+1;
12                 else
13                     end = mid-1;
14             }
15             else {  // left half sorted
16                 if(target>=A[start] && target<A[mid])
17                     end = mid-1;
18                 else
19                     start = mid+1;
20             }
21         }
22         return -1;
23     }
24 };

解法二和解法三参考了：http://bangbingsyb.blogspot.com/2014/11/leetcode-search-in-rotated-sorted-array.html

思路如下：

题目一看就知道是binary search。所以关键点在于每次要能判断出target位于左半还是右半序列。解这题得先在纸上写几个rotated sorted array的例子出来找下规律。Rotated sorted array根据旋转得多少有两种情况：

原数组：0 1 2 4 5 6 7
情况1： 6 7 0 1 2 4 5 起始元素0在中间元素的左边
情况2： 2 4 5 6 7 0 1 起始元素0在中间元素的右边

两种情况都有半边是完全sorted的。根据这半边，当target != A[mid]时，可以分情况判断：

当A[mid] < A[end] < A[start]：情况1，右半序列A[mid+1 : end] sorted
A[mid] < target <= A[end], 右半序列，否则为左半序列。

当A[mid] > A[start] > A[end]：情况2，左半序列A[start : mid-1] sorted
A[start] <= target < A[mid], 左半序列，否则为右半序列

最后总结出：
A[mid] = target, 返回mid，否则

(1) A[mid] < A[end]: A[mid+1 : end] sorted
A[mid] < target <= A[end] 右半，否则左半。

(2) A[mid] > A[end] : A[start : mid-1] sorted
A[start] <= target < A[mid] 左半，否则右半。

转载于:https://www.cnblogs.com/abc-begin/p/7552275.html

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